Question

Nitrogen dioxide \(\left(\mathrm{NO}_{2}\right)\) cannot be obtained in a pure form in the gas phase because it exists as a mixture of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\). At \(25^{\circ} \mathrm{C}\) and \(0.98 \mathrm{~atm}\), the density of this gas mixture is \(2.7 \mathrm{~g} / \mathrm{L}\). What is the partial pressure of each gas?

Short answer

The partial pressures are 0.519 atm for \(NO_2\) and 0.461 atm for \(N_2O_4\).

Step-by-step solution

Calculate Molar Mass of the Mixture

With the density \(d = 2.7 \, \mathrm{g/L}\) and pressure \(P = 0.98 \, \mathrm{atm}\), we use the ideal gas law \(PV = nRT\) rearranged to find molar mass \(M = \frac{dRT}{P}\), where \(R = 0.0821 \, \mathrm{L} \,\mathrm{atm} / \mathrm{K} \, \mathrm{mol}\) and \(T = 298 \, \mathrm{K}\). Substituting these into the formula gives us: \[M = \frac{(2.7 \, \mathrm{g/L}) \times (0.0821 \, \mathrm{L} \, \mathrm{atm}/\mathrm{K} \, \mathrm{mol}) \times 298 \mathrm{K}}{0.98 \, \mathrm{atm}} = 67.6 \, \mathrm{g/mol}\]

Calculate Molar Masses of Components

Calculate the molar masses of \,\(\mathrm{NO}_{2}\,\) and \,\(\mathrm{N}_{2} \mathrm{O}_{4}\,\): \[\mathrm{M(NO_{2}) = 14 + 2 \times 16 = 46 \,\, \mathrm{g/mol}}\] \[ \mathrm{M(N_{2}O_{4}) = 2 \times 14 + 4 \times 16 = 92 \,\, \mathrm{g/mol}}\]

Set up Equation for Mixture's Molar Mass

Let \,\(x\,\) be the mole fraction of \,\(\mathrm{NO}_{2}\,\), then \,\((1-x)\,\) is the mole fraction of \,\(\mathrm{N}_{2} \mathrm{O}_{4}\,\). The equation for molar mass of the mixture becomes: \[67.6 = 46x + 92(1-x)\]

Solve for Mole Fraction of Each Gas

Solve the equation from the previous step: \[67.6 = 46x + 92 - 92x \ 67.6 - 92 = 46x - 92x \ -24.4 = -46x \ x = \frac{24.4}{46} = 0.530.\] Thus, \,\(x \, \approx 0.53\,\) for \,\(\mathrm{NO}_{2}\,\) and \,\(1-x = 0.47\,\) for \,\(\mathrm{N}_{2} \mathrm{O}_{4}\,\).

Calculate Partial Pressures

Since the partial pressure of a gas is its mole fraction times total pressure: \[P_{\mathrm{NO}_{2}} = 0.53 \times 0.98 = 0.5194 \, \mathrm{atm}\] \[P_{\mathrm{N}_{2} O_{4}} = 0.47 \times 0.98 = 0.4606 \, \mathrm{atm}\]

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental principle used for understanding the properties of gases. It relates the pressure, volume, temperature, and amount of a gas. The equation is expressed as:\[ PV = nRT \]- **P** is the pressure of the gas.- **V** is the volume.- **n** is the number of moles of gas.- **R** is the ideal gas constant, which is 0.0821 L·atm/(K·mol).- **T** is the temperature in Kelvin.
We can rearrange this equation to calculate the molar mass of a gas mixture when its density, temperature, and pressure are known. The rearranged formula looks like:\[ M = \frac{dRT}{P} \]where **M** is the molar mass and **d** is the density of the gas mixture.
Understanding and applying this formula is crucial when dealing with gas mixtures, as it allows us to find important properties like density, pressure, and molar mass.

Mole Fraction

The mole fraction is a measure of the concentration of a particular component in a mixture. It represents how many moles of one component are present compared to the total number of moles in the mixture. The formula for the mole fraction of a component, say \( x \) for nitrogen dioxide \( \text{NO}_2 \), is expressed as:\[ x = \frac{n_{ ext{component}}}{n_{ ext{total}}} \]- **\( n_{\text{component}} \)** is the number of moles of the component.- **\( n_{\text{total}} \)** represents the total moles of all components in the mixture.
This concept is vital when solving problems that require the use of partial pressures in gas mixtures. Knowing the mole fraction of each gas allows you to easily calculate the partial pressures within a mixture, which is a key step in many chemical and physical processes involving gases.

Partial Pressure

Each gas in a mixture exerts its own pressure, known as its partial pressure. The total pressure of the gas mixture is the sum of all individual partial pressures. The partial pressure of a particular gas can be computed using its mole fraction and the total pressure of the mixture:\[ P_{ ext{component}} = x_{ ext{component}} \times P_{ ext{total}} \]- **\( P_{\text{component}} \)** is the partial pressure of the gas.- **\( x_{\text{component}} \)** is the mole fraction of the gas.- **\( P_{\text{total}} \)** is the total pressure of the gas mixture.
For example, in a mixture of nitrogen dioxide \( \text{NO}_2 \) and dinitrogen tetroxide \( \text{N}_2\text{O}_4 \), determining their mole fractions using the relationship based on molar masses allows you to find their respective partial pressures. This method is particularly useful in scenarios where each gas contributes differently to a mixture's properties.

Molar Mass

The molar mass is a critical concept that is defined as the mass of one mole of a substance expressed in grams per mole (g/mol). To calculate it, you need to add together the atomic masses of all elements in the molecule. For example:- **Nitrogen Dioxide (NO\(_2\))**: Consists of 1 nitrogen (14 g/mol) and 2 oxygens (2 × 16 g/mol), resulting in a molar mass of 46 g/mol.- **Dinitrogen Tetroxide (N\(_2\)O\(_4\))**: Comprised of 2 nitrogens (2 × 14 g/mol) and 4 oxygens (4 × 16 g/mol), giving a molar mass of 92 g/mol.
In gas mixtures, the effective molar mass can be derived using the weighted average based on mole fractions of the components. This mediating property between molecular weight and composition is critical for calculating the properties of complex gas mixtures and helps to find mole fractions based on known compositions.